PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
2nd Edition
ISBN: 9781285074788
Author: Ball
Publisher: CENGAGE L
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Chapter 8, Problem 8.35E
Interpretation Introduction

(a)

Interpretation:

The ratio of molarity of the ions in the given concentration cells required to have E equal to 0.050V is to be calculated.

Concept introduction:

The type of electrochemical cells which is comprised of two same electrodes but different concentrations or pressures is known as concentration cell.

Expert Solution
Check Mark

Answer to Problem 8.35E

The ratio of [Fe2+]prod[Fe2+]react for the given concentration cell is 0.020.

Explanation of Solution

The given concentration cell is,

Fe2+[react]Fe2+[prod]

For a concentration cell, standard electrode potential E° is zero. The given electrode potential is 0.050V.

The electrode potential at the given concentration can be calculated by using the Nernst equation as shown below.

E=E°RTnFlnQ

Where,

n is the number of electrons

T is the temperature

Q is the reaction quotient

F is the Faraday constant.

E° is the standard electrode potential.

E is the electrode potential at given conditions.

Substitute the values of given potential, standard electrode potential and concentration of the species in the formula.

0.050=0(8.314JKmol)(298K)(2mole)(96485Cmole)ln[Fe2+]prod[Fe2+]react0.050=2477.572192970ln[Fe2+]prod[Fe2+]react0.050=0.0128ln[Fe2+]prod[Fe2+]react0.0500.0128=ln[Fe2+]prod[Fe2+]react

Solve for the value of ratio of [Fe2+]prod[Fe2+]react as shown below.

ln[Fe2+]prod[Fe2+]react=3.91[Fe2+]prod[Fe2+]react=0.020

Thus, the ratio of [Fe2+]prod[Fe2+]react for the given concentration cell is 0.020.

Conclusion

The ratio of [Fe2+]prod[Fe2+]react for the given concentration cell is 0.020.

Interpretation Introduction

(b)

Interpretation:

The ratio of molarity of the ions in the given concentration cells required to have E equal to 0.050V is to be determined.

Concept introduction:

The type of electrochemical cells which is comprised of two same electrodes but different concentrations or pressures is known as concentration cell.

Expert Solution
Check Mark

Answer to Problem 8.35E

The ratio of [Fe3+]prod[Fe3+]react for the given concentration cell is 0.0029.

Explanation of Solution

The given concentration cell is,

Fe3+[reacts]Fe3+[prods]

For a concentration cell standard electrode potential E° is zero. The given electrode potential is 0.050V.

The electrode potential at the given concentration can be calculated by using the Nernst equation as shown below.

E=E°RTnFlnQ

Where,

n is the number of electrons

T is the temperature

Q is the reaction quotient

F is the Faraday constant.

E° is the standard electrode potential.

E is the electrode potential at given conditions.

Substitute the values of given potential, standard electrode potential and concentration of the species in the formula.

0.050=0(8.314JKmol)(298K)(3mole)(96485Cmole)ln[Fe3+]prod[Fe3+]react0.050=2477.572289455ln[Fe3+]prod[Fe3+]react0.050=0.00856ln[Fe3+]prod[Fe3+]react0.0500.00856=ln[Fe3+]prod[Fe3+]react

Solve for the value of ratio of [Fe3+]prod[Fe3+]react as shown below.

ln[Fe3+]prod[Fe3+]react=5.84[Fe3+]prod[Fe3+]react=0.0029

Thus, the ratio of [Fe3+]prod[Fe3+]react for the given concentration cell is 0.0029.

Conclusion

The ratio of [Fe3+]prod[Fe3+]react for the given concentration cell is 0.0029.

Interpretation Introduction

(c)

Interpretation:

The ratio of molarity of the ions in the given concentration cells required to have E equal to 0.050V is to be determined.

Concept introduction:

The type of electrochemical cells which is comprised of two same electrodes but different concentrations or pressures is known as concentration cell.

Expert Solution
Check Mark

Answer to Problem 8.35E

The ratio of [Co2+]prod[Co2+]react for the given concentration cell is 0.020.

Explanation of Solution

The given concentration cell is,

Co2+[reacts]Co2+[prods]

For a concentration cell standard electrode potential E° is zero. The given electrode potential is 0.050V.

The electrode potential at the given concentration can be calculated by using the Nernst equation as shown below.

E=E°RTnFlnQ

Where,

n is the number of electrons

T is the temperature

Q is the reaction quotient

F is the Faraday constant.

E° is the standard electrode potential.

E is the electrode potential at given conditions.

Substitute the values of given potential, standard electrode potential and concentration of the species in the formula.

0.050=0(8.314JKmol)(298K)(2mole)(96485Cmole)ln[Co2+]prod[Co2+]react0.050=2477.572192970ln[Co2+]prod[Co2+]react0.050=0.0128ln[Co2+]prod[Co2+]react0.0500.0128=ln[Co2+]prod[Co2+]react

Solve for the value of ratio of [Co2+]prod[Co2+]react as shown below.

ln[Co2+]prod[Co2+]react=3.91[Co2+]prod[Co2+]react=0.020

Thus, the ratio of [Co2+]prod[Co2+]react for the given concentration cell is 0.020.

Conclusion

The ratio of [Co2+]prod[Co2+]react for the given concentration cell is 0.020.

Interpretation Introduction

(d)

Interpretation:

The answers are to be compared and similarities or differences are to be stated.

Concept introduction:

The type of electrochemical cells which is comprised of two same electrodes but different concentrations or pressures is known as concentrations cells.

Expert Solution
Check Mark

Answer to Problem 8.35E

The ratio of molarities is equal in case of [Co2+]prod[Co2+]react and [Fe2+]prod[Fe2+]react as the number of electron change involved in these cells are same.

Explanation of Solution

The ratio of molarities is equal in case of [Co2+]prod[Co2+]react and [Fe2+]prod[Fe2+]react as the number of electron change involved in these cells are the same. The concentration cells do not depend on the chemical identities of the half-cells involved.

Conclusion

The ratio of molarities is equal in case of [Co2+]prod[Co2+]react and [Fe2+]prod[Fe2+]react as the number of electron change involved in these cells are same.

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Chapter 8 Solutions

PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.

Ch. 8 - Prob. 8.11ECh. 8 - Prob. 8.12ECh. 8 - 8.13. Is the disproportionation reaction...Ch. 8 - Prob. 8.14ECh. 8 - Prob. 8.15ECh. 8 - Prob. 8.16ECh. 8 - Prob. 8.17ECh. 8 - 8.18. Determine and for each of the following...Ch. 8 - Prob. 8.19ECh. 8 - Prob. 8.20ECh. 8 - Prob. 8.21ECh. 8 - Prob. 8.22ECh. 8 - Prob. 8.23ECh. 8 - Prob. 8.24ECh. 8 - Prob. 8.25ECh. 8 - Prob. 8.26ECh. 8 - Prob. 8.27ECh. 8 - What is the Zn2+:Cu2+ ratio on a Daniell cell that...Ch. 8 - Prob. 8.29ECh. 8 - Determine the voltage of this reaction with the...Ch. 8 - The thermite reaction can act as the basis of an...Ch. 8 - A concentration cell has different concentrations...Ch. 8 - Prob. 8.34ECh. 8 - Prob. 8.35ECh. 8 - a What is the equilibrium constant for the...Ch. 8 - Prob. 8.37ECh. 8 - Prob. 8.38ECh. 8 - Prob. 8.39ECh. 8 - Prob. 8.40ECh. 8 - Prob. 8.41ECh. 8 - Consider the following formation reaction for HI:...Ch. 8 - Prob. 8.43ECh. 8 - 8.44. Determine an expression for , the change in...Ch. 8 - Prob. 8.45ECh. 8 - Prob. 8.46ECh. 8 - Determine the equilibrium constant for the...Ch. 8 - Prob. 8.48ECh. 8 - Prob. 8.49ECh. 8 - What is the solubility product constant of Hg2Cl2,...Ch. 8 - Prob. 8.51ECh. 8 - Prob. 8.52ECh. 8 - Prob. 8.53ECh. 8 - Prob. 8.54ECh. 8 - Prob. 8.55ECh. 8 - Prob. 8.56ECh. 8 - Prob. 8.57ECh. 8 - Show that a can be written as n+mnn+n+nn, where m...Ch. 8 - Prob. 8.59ECh. 8 - Prob. 8.60ECh. 8 - What molality of NaCl is necessary to have the...Ch. 8 - Prob. 8.62ECh. 8 - Prob. 8.63ECh. 8 - Calculate the molar enthalpy of formation of I(aq)...Ch. 8 - Prob. 8.65ECh. 8 - Hydrofluoric acid, HF(aq), is a weak acid that is...Ch. 8 - Prob. 8.68ECh. 8 - Prob. 8.69ECh. 8 - Prob. 8.70ECh. 8 - Prob. 8.71ECh. 8 - Prob. 8.72ECh. 8 - The mean activity coefficient for an aqueous...Ch. 8 - Human blood plasma is approximately 0.9NaCl. What...Ch. 8 - Under what conditions does the extended...Ch. 8 - Prob. 8.76ECh. 8 - Approximate the expected voltage for the following...Ch. 8 - Prob. 8.78ECh. 8 - Prob. 8.79ECh. 8 - Prob. 8.80ECh. 8 - a The salt NaNO3 can be thought of as...Ch. 8 - Prob. 8.82ECh. 8 - What is the estimated velocity for Cu2+ ions...Ch. 8 - Prob. 8.84ECh. 8 - Prob. 8.85ECh. 8 - Prob. 8.86ECh. 8 - Calculate a the solubility product constant for...
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